- #1
humphreybogart
- 22
- 1
I am working my way through a textbook, and whenever this equation is solved (integrated), the answer is given as:
u = f(x) + f(y)
I don't understand it. If I integrate it once (with respect to y, say), then I obtain:
∂u/∂x = f(x) -----eq.1
If I integrate again (this time with respect to x), then I obtain:
u = xf(x) + f(y)
I know that this can't be correct because the mixed derivative theorem says that if I went the other way (integrating with respect to x and then y), I should get the same answer. But I can't see how integrating eq.1 doesn't produce and 'x' infront of the arbitrary function.
u = f(x) + f(y)
I don't understand it. If I integrate it once (with respect to y, say), then I obtain:
∂u/∂x = f(x) -----eq.1
If I integrate again (this time with respect to x), then I obtain:
u = xf(x) + f(y)
I know that this can't be correct because the mixed derivative theorem says that if I went the other way (integrating with respect to x and then y), I should get the same answer. But I can't see how integrating eq.1 doesn't produce and 'x' infront of the arbitrary function.